I am wondering how to generate a random double number inside the entire double range in Python, so that the order of magnitude of the number (math.log10) has an equal chance to be any of the possible magnitudes (range(-307, 309)).

I tried the following code but it does not give intended result:

from random import random, uniform
from collections import Counter
import math
import sys

random_numbers = [random()/random() for i in range(2048)]

count = Counter([int(math.log10(i)) for i in random_numbers])

print(count.most_common())

uniform_numbers = [uniform(sys.float_info.min, sys.float_info.max) for i in range(2048)]

count1 = Counter([int(math.log10(i)) for i in uniform_numbers])

print(count1.most_common())

[(0, 1862), (1, 93), (-1, 68), (2, 14), (-2, 10), (3, 1)] 
[(307, 1055), (308, 888), (306, 95), (305, 10)]

CodePudding user response：

Because logarithm and exponential are inverse functions of eachother, you can do that by generating a number along a uniform distribution, and then compute its exponential

import numpy as np
N = 100
x = 10 ** np.random.uniform(low=-307, high=309, size=N)

keep in mind that the resulting distribution will not be uniform (because it is "distorted" by the exponential). Very often, the sample will be very close to 0 or very close to infinity.